Question

Consider three gas connected gas containers as shown in the figure below.

Consider three gas connected gas containers as shown in the figure below.

CO2 (g) P = 2.13, V = 1.50L

H2(g) P = 0.861 atm, V = 1.00 L

Ar (g) p = 1.15atm, V = 2.00L

   T = 298K

Assume, that you open the two stopcock so that the gases can flow freely.

Calculate the change of the chemical potential of the entire system due to mixing the gases.

Using the change of the chemical potential of gases show that the mixing of three gases is spontaneous

Determine the value of D_mixH .

_mix = ______________________

_mixH = _____________________

Answer 1

The chemical potential of a gas is given as

μ_i = μ_i⁰ + RTlnP_i where μ_i is the chemical potential of the i-th component in the system at pressure P_i and μ_i⁰ is the standard chemical potential of the i-th component at 1 bar pressure.

Let us calculate the number of moles of each component as

Component	Initial pressure (atm)	Initial volume (L)	Number of moles = PV/RT
CO2 = 1	2.13	1.50	(2.13)(1.50)/RT = 3.195/RT
H2 = 2	0.861	1.00	(0.861)(1.00)/RT = 0.861/RT
Ar = 3	1.15	2.00	(1.15)(2.00)/RT = 2.30/RT

Since chemical potential is an additive property, the Gibb’s free energy of the system before mixing the gases is

G_i = n₁μ₁ + n₂μ₂ + n₃μ₃ = [3.195/RT{μ₁⁰ + RTln(2.13)} + 0.861/RT{μ₂⁰ + RTln(0.861)} + 2.30/RT{μ₃⁰ + RTln(1.15)}]

or, G_i = {3.195ln(2.13) + 0.861ln(0.861) + 2.30ln(1.15)} + 1/RT(3.195μ₁ + 0.861μ₂ + 2.30μ₃) ……(1)

When the partition is removed, the total volume of the system is 4.5 L and the partial pressures of components 1, 2 and 3 are

P₁’= (3.195/RT)(RT/4.00) = 0.79875 atm (since we assume ideal behaviour and P’= nRT/V where V is the total volume and n is the number of mole of the said component)

P₂’= (0.861/RT)(RT/4.00) = 0.21525 atm and

P₃’= (2.30/RT)(RT/4.00) = 0.575 atm.

As before, the Gibb’s free energy after mixing is

G_f = n₁μ₁’+ n₂μ₂’+ n₃μ₃’

Hence, G_f = [(3.195/RT){μ₁⁰ + RTln(0.79875)} + (0.861/RT){μ₂⁰ + RTln(0.21525)} + (2.30/RT){μ₃⁰ + RTln(0.575)}]

or, G_f = {3.195ln(0.79875) + 0.861ln(0.21525) + 2.30ln(0.575)} + 1/RT(3.195μ₁⁰ + 0.861μ₂⁰ + 2.30μ₃⁰) …….(2)

The Gibb’s free energy of mixing is

ΔG_m = G_f – G_i = {3.195ln(0.79875) + 0.861ln(0.21525) + 2.30ln(0.575)} – {3.195ln(2.13) + 0861ln(0.861) + 2.30ln(1.15)}

or, ΔG_m = 3.195ln(0.79875/2.13) + 0.861ln(0.21525/0.861) + 2.30ln(0.575/1.15) = (- 3.134) + (-1.193) + (-1.594) = -5.921

The free energy change of mixing is -5.92 kJ (Gibb’s free energy change is always expressed in kJ)

Since, ΔG_m < 0, the mixing process is spontaneous.

The entropy of mixing, ΔS_m = -ΔG_m/T where T is the absolute temperature.

The enthalpy of mixing can be calculated using the relation between G, H and S.

ΔG_m = ΔH_m – T.ΔS_m ==è ΔH_m = ΔG_m + T.ΔS_m = ΔG_m + T. (-ΔG_m/T) = 0